Split - More Integrals

[1John5vs7]

Two more that I just can't figure out how to integrate:


I wish I knew how to write the math out fancy-style like you guys do. I don't see anything in the buttons for forum posts, so I assume it's some kind of in-line code or something. Anyway, here are the two giving me fits:

1.) Integral[ 1 / (x^2 -4) ]dx
U substitution doesn't look like it would work, and neither would integration by part really....I don't know of any rules for x^2 - a^2 unless it's under a radical...although I guess if (Root(x^2-a^2)) = a*tan(theta) that is a possibility... is that what I ought to go with you think?

Edit: I tried taking a look at the solution given for the problem, but I don't know how to take the derivative I'm sad to say. The answer is:

1/4*(ln|(x-2)/(x+2)|+c.

I know the derivative of ln|x+2| is dx/(x+2), but what do I do with the (x-2)?

2.) Definite integral from 1 to 2 [ (Root (u^2 - 1))]du
This one I recognize is an improper integral because of the discontinuity at u=1. So I'm rewritten it:

lim as u-->1+ Def Integral from 1 to 2 [Root (u^2-1))]du and from there I've tried a few different approaches. My understanding is (Root(u^2-1)) = 1*tan(theta), so I've set this equal to:

Integral[ 1*tan(theta)]d(theta) which, according to my textbook, has a common integral of ln|sec(theta)|.

I don't know where to go from there though. I *do* have the answer for it and will try taking the derivative of it to see what fruit that bears. Same is true for the first one in this reply. Man...is Calc II supposed to make your eyes water, your head throb, and your desire to live plummet? Criminey, I've been at it for the last 4 days doing nothing but this stuff. I've got 13 more class meetings to go until it's over, so I guess I'll make it. Wish me luck!

Edit: Duh, I just realized that the answer for the second problem I have listed here is a number value because it's a definite integral. As such, I can't use it to take a derivative and try to trace back to the original integral. Que sera sera.

 

[stapel]

I wish I knew how to write the math out fancy-style like you guys do.

To learn, try doing a "reply with quote". The coding will be included in the reply's text, especially if you click the far-left button in the top row of formatting options (the "A/A" button). Anything you see inside "tex" tags is probably the pretty math stuff.

1.) Integral[ 1 / (x^2 -4) ]dx

The denominator factors, so try partial fractions:

. . . . .\(\displaystyle x^2\, -\, 4\, =\, (x\, -\, 2)(x\, +\, 2)\)

Doing the algebra, you'll find:

. . . . .\(\displaystyle \dfrac{1}{x^2\, -\, 4}\, =\, \dfrac{1}{4(x\, -\, 2)}\, -\, \dfrac{1}{4(x\, +\, 2)}\)

Can you take it from there?

2.) Definite integral from 1 to 2 [ (Root (u^2 - 1))]du
This one I recognize is an improper integral because of the discontinuity at u=1.

Unless the radical is in the denominator (and I'm not seeing any fractions here), the integrand is well-defined over the interval. Yes, it's equal to zero at the left-hand endpoint, but that's fine. Just do the integration. (By the way, this old thread may be helpful.)

 

[1John5vs7]

To learn, try doing a "reply with quote". The coding will be included in the reply's text, especially if you click the far-left button in the top row of formatting options (the "A/A" button). Anything you see inside "tex" tags is probably the pretty math stuff.


The denominator factors, so try partial fractions:

. . . . .\(\displaystyle x^2\, -\, 4\, =\, (x\, -\, 2)(x\, +\, 2)\)

Doing the algebra, you'll find:

. . . . .\(\displaystyle \dfrac{1}{x^2\, -\, 4}\, =\, \dfrac{1}{4(x\, -\, 2)}\, -\, \dfrac{1}{4(x\, +\, 2)}\)

Can you take it from there?


Unless the radical is in the denominator (and I'm not seeing any fractions here), the integrand is well-defined over the interval. Yes, it's equal to zero at the left-hand endpoint, but that's fine. Just do the integration. (By the way, this old thread may be helpful.)

Arg! Of course! Partial fractions. Thanks, you've given me the clue I needed. You're a brick!!!

Also, I don't know what I was on about with the discontinuity. Fatigue setting in. Brain must sleep now. But thanks for pointing that out.... I feel like a goof now

I am looking at the problem that I thought had the discontinuity and still can't figure out how to integrate. Once I get to tan(theta)d(theta), that integrates to ln|sec(theta)| but then I have to change the limits of the interval to terms of pi, right? I guess I don't know how to do that since I don't have a u-substitution equation to plug back in to, and I don't know how to convert sec(theta) back into terms of x either so I feel a little sunk.

 

[Steven G]

Two more that I just can't figure out how to integrate:


I wish I knew how to write the math out fancy-style like you guys do. I don't see anything in the buttons for forum posts, so I assume it's some kind of in-line code or something. Anyway, here are the two giving me fits:

1.) Integral[ 1 / (x^2 -4) ]dx equal to:

Stapel's method should work fine. My concern is that it seems based on the problems you submitted that you are working on trig substitution problems. So lets try solving it that way. Who knows, it may not work but that is the thing with substitution. As a student I was upset with my teacher for not ever making an incorrect substitution so that the students (myself included) could see why a sub did not work.

Three different types.
1) you have a^2+x^2 to any power in the integral. Place sqrt(a^2+u^2) on the hypotenuse and a and u on the legs. Then make a trig subt that will yield u/a.

2) you have a^2-u^2 to any power. Place a on the hypotenuse and u on one of the legs. Put sqrt(a^2-u^2) on the other leg (please confirm this last side by using the Pythagorean theorem) Now make a trig sub that yields u/a (draw the triangle, call the angle u and the rest should follow).

3) you have u^2-a^2 to any power in the integral. Place u on the hypotenuse and a on one of the legs. The other leg gets sqrt(u^2-a^2)-verify this! Now make a trig sub that yields u/a.

DO NOT memorize the above. It needs to be the obvious substitutions in your mind.

Note that in all 3 cases you place the sqrt of the positive part on the hypotenuse, on one leg you place the sqrt of the negative part (if there is one).

If you have any question then get back to us.